October 6th, 2012, 10:17 AM  #1 
Newbie Joined: Jan 2012 Posts: 9 Thanks: 0  Irrational numbers
How can I prove that sum of this twon numbers in this link is irrational Link http://imageshack.us/photo/myimages/109/scanyr.jpg/ Thank you 
October 6th, 2012, 11:17 AM  #2 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233  Re: Irrational numbers
We know that is not an integer, then you can use that result to show is irrational by way of contradiction. Suppose for integer Now look at this very carefully... a and b are integers. We know because it was in the original denominator, we know because the original sum is not zero. is an integer, has no chance of being an integer because the expression inside the parenthesis is NOT an integer since is NOT an integer. Conclusion: We assumed there were integers a,b to make the sum rational, we found this to be impossible therefore the sum is irrational. I'm sure others will post better proofs, wait for them. 
October 7th, 2012, 07:23 PM  #3  
Senior Member Joined: Nov 2011 Posts: 595 Thanks: 16  Re: Irrational numbers Quote:
 
October 7th, 2012, 08:46 PM  #4 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233  Re: Irrational numbers
That's a good point Dougy, i'm going to have to rethink on it. If this 'thing'... ... is the cube root of an integer then my argument is useless. Maybe i can fix my argument? Those radicals are going to prevent the expanded expression on the right from being an integer since a, b can't be zero. There is another possibility... For nonzero b this has no real number solutions. So, have i fixed my argument? Let me restate my argument (and change the wording). For nonzero integer a and b , cannot be an integer. 
October 7th, 2012, 09:01 PM  #5 
Senior Member Joined: Nov 2011 Posts: 595 Thanks: 16  Re: Irrational numbers
Yes I think that (almost) finishes the proof now! However I think you did not treat the most general case. You want to prove that can not be an integer/ Now you say that is equivalent to prove that is not an integer. Since, as you said, is an integer, that means this is equivalent to show that can not be an integer. Now I don't understand why you put this thing to be equal to zero? It could be equal to any integer no? In any case, the proof is saved, because if we assume there exists an integer P such that means means which means that is rational, which is absurd... edit: here P is a negative integerof course..so it is more like a relative number than an integer, question of wording! 
October 7th, 2012, 09:10 PM  #6 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233  Re: Irrational numbers
You are right again Dougy, in my mind i wanted to make the radicals 'go away' that's why i set it equal to zero. You are correct, I would have to rule out every integer. You saved the proof very nicely. I was not thorough... 

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